How to sort the output obtained with grep -c?

Question

I use the following 'grep' command to get the count of the string alert in each of my files at the given path:

grep 'alert' -F /usr/local/snort/rules/* -c

How do I sort the resulting output in desired order- say ascending order, descending order, ordered by name, etc. An answer specific to these cases is sufficient.

You may freely suggest a command other than grep as well.

Possible duplicate of [Unix Sort with Tab Delimiter](http://stackoverflow.com/questions/1037365/unix-sort-with-tab-delimiter) with column == 2 and separator == `:` — Ciro Santilli OurBigBook.com, Nov 22 '15 at 10:28

score 31 · Accepted Answer · answered Sep 04 '12 at 08:17

Pipe it into sort. Assuming your filenames have no colons, use the "-t" option to specify the colon as field saparator. Use -n for numerical sorting.

Example:

grep 'alert' -F /usr/local/snort/rules/* -c | sort -t: -n -k2

should split lines into fields separated by ":", use the second field for sorting, and treat this as numbers (so 21 is actually later than 3).

How to sort the output obtained with grep -c?

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